Two negative integers are five minutes apart on the number line, and their product is 126. What is the sum of the two integers?

1 answer:

7 0

First I will assume that this problem can be reworded as follows: "<span>Two negative integers are five units apart on the number line."

Let the first (the smaller) negative integer be x and the second y. Note that y, being bigger than x, can be represented by y=x+5.

First find the 2 integers. Their product is 126, so x(x+5)=126.
Expanding, x^2 + 5x - 126 = 0. Using the quadratic formula to solve this,

-5 plus or minus sqrt(529) -5 plus or minus 23
x = --------------------------------------- = -----------------------------
2 2
-5 -23
We are interested only in negative integers, so take x = ------------
2

Then x = -28/2, or x = -14. Then y is x+5, or -14+5 = -9.

Check: does the product of -9 and -14 equal 126, as it must? YES.</span>

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D = L/k

Step-by-step explanation:

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dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

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1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$